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Ans:
 
Ans:
<math> a_0= average of x(t) = /frac {1x2}{4}=/frac{1}{2};
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<math>a_0 = </math>average of x(t) = <math>\frac{1x2}{4}=\frac{1}{2}; </math>
  
 
T=4, <math>\omega _o = \frac{2 \pi} {4} = \frac{\pi}{2}</math>  
 
T=4, <math>\omega _o = \frac{2 \pi} {4} = \frac{\pi}{2}</math>  
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<math>a_k=\frac{1}{4} \left.\frac{e^{-jk\frac{\pi}{2}t}}{-jk\frac{\pi}{2}}\right|_{-1}^{1}</math>
 
<math>a_k=\frac{1}{4} \left.\frac{e^{-jk\frac{\pi}{2}t}}{-jk\frac{\pi}{2}}\right|_{-1}^{1}</math>
  
<math>a_k=\frac{1}{4} (\frac{e^{-jk\frac{\pi}{2}t}}{-jk\frac{\pi}{2}}-\frac{e^{jk\frac{\pi}{2}t}}{jk\frac{\pi}{2}})<\math>
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<math>a_k=\frac{1}{4} (\frac{e^{-jk\frac{\pi}{2}t}}{-jk\frac{\pi}{2}}-\frac{e^{jk\frac{\pi}{2}t}}{jk\frac{\pi}{2}})</math>
  
<math>a_k=\frac{1}{2jk\pi}(e^{jk\frac{\pi}{2}}-e^{-jk\frac{\pi}{2}})<\math>
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<math>a_k=\frac{1}{2jk\pi}(e^{jk\frac{\pi}{2}}-e^{-jk\frac{\pi}{2}})</math>
  
<math>a_k=\frac{sin(k\frac{\pi}{2})}{k\pi}<\math>
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<math>a_k=\frac{sin(k\frac{\pi}{2})}{k\pi}</math>

Latest revision as of 08:12, 15 October 2008

Question 4 Compute the coefficients $ a_k $ of the Fourier series of the signal $ x(t) $ periodic with period $ T=4 $ defined by $ \,x(t)=\left\{\begin{array}{cc} 0, & -2<t<-1 \\ 1, & -1\leq t\leq 1 \\ 0, & 1<t\leq 2 \end{array} \right. \, $

(Simplify your answer as much as possible.)



Ans:

$ a_0 = $average of x(t) = $ \frac{1x2}{4}=\frac{1}{2}; $

T=4, $ \omega _o = \frac{2 \pi} {4} = \frac{\pi}{2} $

for k<>0,

$ a_k=\frac{1}{T}\int_{0}^{T}x(t)e^{-jk\frac{2\pi}{T}t}dt $

$ a_k=\frac{1}{4}\int_{-2}^{2}x(t)e^{-jk\frac{\pi}{2}t}dt $

$ a_k=\frac{1}{4}\int_{-1}^{1}x(t)e^{-jk\frac{\pi}{2}t}dt $

$ a_k=\frac{1}{4} \left.\frac{e^{-jk\frac{\pi}{2}t}}{-jk\frac{\pi}{2}}\right|_{-1}^{1} $

$ a_k=\frac{1}{4} (\frac{e^{-jk\frac{\pi}{2}t}}{-jk\frac{\pi}{2}}-\frac{e^{jk\frac{\pi}{2}t}}{jk\frac{\pi}{2}}) $

$ a_k=\frac{1}{2jk\pi}(e^{jk\frac{\pi}{2}}-e^{-jk\frac{\pi}{2}}) $

$ a_k=\frac{sin(k\frac{\pi}{2})}{k\pi} $

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